PSTAT 5E Practice Homework 3 Solutions

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1. PSTAT 5E. Practice Homework 3. Solutions. 3.12. Each simple event is equally likely, with probability 1/5. a. S: E1, E2, E3, E4, E5. P(S) = 5/5 =1 b. A: E1, E3.

3.12 Each simple event is equally likely, with probability 1/5. a. S: E1, E2, E3, E4, E5

P(S) = 5/5 =1

b. A: E1, E3

P(A) = 2/5

c. B: E1, E2, E4, E5

P(B) = 4/5

d. C: E3, E4

P(C) = 2/5

e. Α : E2, E4, E5

P( Α ) = 3/5

f. Β : E3

P( Β ) = 1/5

g. AB: E1

P(AB) = 1/5

h. AC: E3

P(AC) = 1/5

i. A B: E3

P(A B) = 1/4

j. A

P(A B) = 1

B: S

k. A C: E1, E3, E4

P(A C) = 3/5

l. A B: E3

P(A B) = 1/2

3.13. From Exercise 3.12, P (A and B) = 1/5, P (A B) = 1/4, and P (A) = 2/5. Since P (A and B) A and B are not mutually exclusive. Since P (A B) P (A), A and B are not independent.

0,

3.18 Define the following events: L: company maintains a parental leave program S: company pays salary H: company pays health care It is given that P (L) = .27; P (S L) = 1/3; P (H L) = ¾

1

1 a. P (L and S) = P (L)*P (S L) = .27 ( ) = .09 3 1 b. P (L and Η ) = P (L)*P ( Η L) = .27 ( ) = .0675 4

3.22. There are 35 brands from which to choose, each of which is categorized according to quality. 2 a. P (E) = 35 11 2 21 34 b. P ( at least “good”) = P ( G or V or E ) = P (G) + P (V) + P (E) = + + = 35 35 35 35 Note: Since G, V, and E are simple events, they are mutually exclusive. Therefore, the probability of the union is the sum of the probabilities of the simple events. 11 + 1 + 0 12 c. P ( not V or E ) = P ( G or F or P ) = = 35 35 d. Define the following events: A: first brand is “very good” B: second brand is “very good” Using the Multiplicative Law of Probability, 21 20 420 P (A and B) = P (A)*P (B A) = = = .3529 35 34 1190 Note: It is also clear from the result above that the events A and B are not independent. 3.24. Use Bayes’ Rule. The denominator given in the formula for P (Si I) is P (I) =

3 i =1

P (Si) P (I | S i ) = .4(.1) + .5(.3) + .1(.2) = .21,

so that P (S1 I) = P (S2 I) = P (S3 I) =

P(S1 )Ρ(Ι S1 ) Ρ (Ι ) P(S 2 )Ρ(Ι S 2 ) Ρ (Ι ) P(S 3 )Ρ(Ι S 3 ) Ρ(Ι)

=

.4(.1) = .1905 .21

=

.5(.3) = .7143 .21

=

.1(.2) = .0952 .21

3.25. This is similar to Exercise 3.24. The denominator given in the formula for P (Si I) is P (I) =

4

P(Si) P(I Si) = .6(.1) + .2(.4) + .2(.3) + .5(.2) = .30

i =1

2

so that P (S1 I) = P (S2 I) =

P(S1 )Ρ(Ι S1 ) Ρ(Ι ) P(S 2 )Ρ(Ι S 2 ) Ρ (Ι )

=

.6(.1) = .2000 .30

P (S3 I) =

=

.2(.4) = .2667 .30

P (S4 I) =

P(S 3 )Ρ(Ι S3 ) Ρ(Ι ) P(S 4 )Ρ(Ι S 4 ) Ρ (Ι )

=

.2(.3) = .2000 .30

=

.5(.2) = .3333 .30

3.26. Define the following events: D: item is defective C: item goes through a complete inspection It is given that P (D) = .1, P(C D) = .6, Ρ(C D) = .2. The probability of interest is P(D C) , the probability that an item is defective given that it goes through a complete inspection. Using Bayes’s Rule

P(D) P(C D)

P(D C) =

P(C)

=

P(C D) P(D) P(C D) P(D) + P(C D) P(D)

=

3.29. a. Since one of the requirements of a probability distribution is that

.6(.1) = .25 .6(.1) + .2(.9)

p ( x) =1, we have x

p(x) = 1 – (.1 + .3 + .4 + .1 + .05 ) = 1 – .95 = .05

b. The probability histogram is shown below 0.5

0.4

p(x)

0.3

0.2

0.1

0 1

2

3

4

5

6

x

3

3.30. a. P [x = 2] = P (2) = .4 and P [x b. P [x

2] = P (2) + P (3) + P (4) + P (5) = .6

3] = P (0) + P (1) + P (2) + P (3) = .9

c. µ = E(x) = x*P(x) = 0(.1) + 1(.3) + 2(.4) + 3(.1) + 4(.05) + 5(.05) = 1.85

Then, the variance is 2 = E[(x – µ )2 ] = (0 - 1.85)2 (.1) + (1-1.85)2 (.3) + 3.31. a. Since 0

P (x)

1 and

+ (5-1.85)2 (.05) = 1.4275

p ( x) = 1, this is a valid probability distribution. x

b. This is not a valid probability distribution, since P (0) is negative. c. This is not a valid probability distribution, since the sum of all the probabilities is not equal to one. 3.34. a. E(x) = x*P(x) = 3(.03) + 4(.05) + + 13(.01) = 7.9 2 2 b. = (x – µ ) p(x) = (3 – 7.9)2 (.3) + (4 – 7.9)2 (.05) + + (13 – 7.9)2 (.01) = 4.73 and = 4.73 = 2.1749 c. Calculate µ ± 2 = 7.9 ± 4.350 or 3.55 to 12.25. Then, referring to the probability distribution of x, P [3.55 x 12.25 ] = P [ 4 x 12 ] = 1 – P (3) – P (13) = 1 – .04 = .96 3.42. Define the following events: A: residents is under 25 years of age C: resident is 25-34

B: resident is over 65 years or older D: resident is 35-44

It is given that P (A) = .15, P (B) = .12, P (C) = .28 and P (D) = .21. Four persons 18 years or older are chosen at random, forming independent events. a. P(AAAA) = [P(A)]4 = (.15)4 = .0005 b. P(BBBB) = [P(B)]4 = (.12)4 = .0002 c. Since the event of interest will occur if either the first, second, third, or fourth person is under 25, and the remaining people are 25 or older, there are four simple events of interest, each of which have probability (.15)(.85)(.85)(.85). Therefore, the probability of interest is 4 P(A) [P( )]3 = 4(.15)(.85)3 = .3685

4

3.45. a. µ = E(x) = x*P(x) = .3 + .6 + .6 + .4 + .25 = 2.15 b. 2 = (x – µ )2 P(x) = (-2.15)2 (.05) + (-1.15)2 (.05) + + (2.85)2 (.05) = 1.5275 c. Calculate = 1.5275 = 1.236. Then µ ± 2 = 2.15 ± 2.472 or -.322 to 4.622. The graph of p(x) with the interval µ ± 2 superimposed is shown below 0.35

0.3

0.25

0.2

P(x) 0.15

0.1

0.05

0 0

1

2

3

4

5

x

5